Introduction

The question of whether it is correct to say that 'if four points are not coplanar, there is one and only one point equidistant from the four points' is a fascinating one in the realm of geometry and topology. This piece will explore this statement, providing a detailed geometric construction to support or refute the claim. We will delve into the underlying principles and provide a rigorous proof.

Understanding Non-Coplanar Points

Let us begin by defining what it means for four points to be non-coplanar. In geometry, a set of points is coplanar if they all lie in the same plane. Conversely, if a set of points cannot lie in a single plane, they are termed non-coplanar. The statement suggests that for any non-coplanar set of four points, there exists a unique point that is equidistant from all four of them. This point is often referred to as the circumcenter in the context of a sphere that passes through all four points.

Geometric Construction and Proof

The construction of the equidistant point involves several steps. We will start by choosing any three non-collinear points, which will form a non-flat triangle. The key idea is that the circumcenter of the triangle formed by these three points, when extended, will intersect a specific plane that is equidistant from all four points.

Step 1: Construct a Triangle and Find the Circumcenter

First, select any three non-collinear points, say (A), (B), and (C). These three points define a unique plane (Pi) and a non-flat triangle (tau). The triangle (tau) has a circumcenter, (S), which is the point where the perpendicular bisectors of its sides intersect. This circumcenter (S) is the center of a circle that passes through all three points (A), (B), and (C).

Step 2: Construct the Perpendicular Bisector to the Third Point

The circumcenter (S) lies on a line (p) that is perpendicular to the plane (Pi) and passes through (S). This line (p) represents the axis of the sphere that can pass through the points (A), (B), and (C).

Step 3: Introduce the Fourth Point and Construct the Bisecting Plane

Now, introduce the fourth point (D), which is non-coplanar with the other three. Construct a plane (b) that is the perpendicular bisector of the line segment (CD). This plane (b) intersects the line (p) at exactly one point, denoted as (O).

Step 4: Prove the Existence of the Equidistant Point

The point (O) is the point we are looking for. It lies on the line (p) and the plane (b). Because (O) lies on the line (p), it is equidistant from the points (A), (B), and (C). Since (O) lies on the plane (b), it is also equidistant from the point (D). Therefore, (O) is the unique point that is equidistant from all four points (A), (B), (C), and (D).

Conclusion

The construction and proof demonstrate that if four points are not coplanar, there indeed exists a unique point (O) that is equidistant from the four points. The point (O) is the intersection of the plane (b), the perpendicular bisector of the line segment from (C) to (D), and the line (p) that is perpendicular to the plane (Pi) and passes through the circumcenter (S). This unique point (O) satisfies the condition of being equidistant from all four points, confirming the statement's correctness.

Further Reading and Exploration

To delve deeper into such geometric problems and understand more problem-solving techniques, I recommend exploring the following resources:

The Quora post on 'Of Planes and Spheres' for a detailed discussion on a related yet slightly different problem. My YouTube channel, ProbLemma, for a rich collection of videos discussing mathematics, physics, and computer science.

By following the steps and constructions outlined here, you can effectively solve similar problems and gain a deeper understanding of geometric concepts.